6.2 Henry’s Law and Dilute Nonelectrolyte Solutions
253
For a dilute solution n 1 is much larger than the other amounts, and the molality is nearly
proportional to the mole fraction:
x i ≈
n i M 1
w 1
m i M 1 (dilute solution)
(6.2-10)
For a dilute solution Henry’s law can be expressed in terms of the molality:
P i k i m i M 1 k
(m)
i m i (dilute solution)
(6.2-11)
where k
(m)
i
k i M 1 is called the molality Henry’s law constant for substance i. For a
sufficiently dilute solution it is independent of the molality but depends on the identities
of all substances present and on the temperature.
E X A M P L E 6.6
a. From the value of k 2 for ethanol (substance 2) in Example 6.4, find the value of k
(m)
2 .
b. Find the vapor pressure of a 0.0500 mol kg −1 solution of ethanol in benzene, assuming
the molality version of Henry’s law to hold.
Solution
a. k
(m)
2 (1.51 × 10 3 torr)(0.07812 kg mol −1 ) 118 torr(mol kg −1 ) −1
b. P 2
118 torr(mol kg −1 ) −1
0.0500 mol kg −1
5.90 torr
For a dilute solution, the chemical potential of a solute can be expressed in terms of
the molality in an equation similar to Eqs. (6.2-2) and (6.1-8). Using Eqs. (6.2-2) and
(6.2-10),
µ i µ
◦(H)
i
+ RT ln(m i M 1 ) (dilute solution)
µ i µ
◦(m)
i
+ RT ln(m i /m ◦ ) (dilute solution)
(6.2-12)
where
µ
◦(m)
i
µ
◦(H)
i
+ RT ln(M 1 m
◦ )
(6.2-13)
and where m ◦ is defined to equal 1 mol kg −1 (exactly).
The quantity µ
◦(m)
i
is the chemical potential of substance i in its molality standard
state. This standard state is component i in a hypothetical solution with m i equal to
m ◦ (exactly 1 mol kg −1 ) and with Henry’s law in the form of Eq. (6.2-11) valid at this
molality. Again we specify a pressure of exactly 1 bar for this standard state. Since the
standard state is a hypothetical solution, the actual 1-molal solution is not required to
obey Henry’s law.
Exercise 6.11
Show that µ
◦(m)
i
is equal to the chemical potential of substance i in the vapor phase at equilibrium
with a 1.000 mol kg −1 solution if Eq. (6.2-11) is valid at this molality.
253
For a dilute solution n 1 is much larger than the other amounts, and the molality is nearly
proportional to the mole fraction:
x i ≈
n i M 1
w 1
m i M 1 (dilute solution)
(6.2-10)
For a dilute solution Henry’s law can be expressed in terms of the molality:
P i k i m i M 1 k
(m)
i m i (dilute solution)
(6.2-11)
where k
(m)
i
k i M 1 is called the molality Henry’s law constant for substance i. For a
sufficiently dilute solution it is independent of the molality but depends on the identities
of all substances present and on the temperature.
E X A M P L E 6.6
a. From the value of k 2 for ethanol (substance 2) in Example 6.4, find the value of k
(m)
2 .
b. Find the vapor pressure of a 0.0500 mol kg −1 solution of ethanol in benzene, assuming
the molality version of Henry’s law to hold.
Solution
a. k
(m)
2 (1.51 × 10 3 torr)(0.07812 kg mol −1 ) 118 torr(mol kg −1 ) −1
b. P 2
118 torr(mol kg −1 ) −1
0.0500 mol kg −1
5.90 torr
For a dilute solution, the chemical potential of a solute can be expressed in terms of
the molality in an equation similar to Eqs. (6.2-2) and (6.1-8). Using Eqs. (6.2-2) and
(6.2-10),
µ i µ
◦(H)
i
+ RT ln(m i M 1 ) (dilute solution)
µ i µ
◦(m)
i
+ RT ln(m i /m ◦ ) (dilute solution)
(6.2-12)
where
µ
◦(m)
i
µ
◦(H)
i
+ RT ln(M 1 m
◦ )
(6.2-13)
and where m ◦ is defined to equal 1 mol kg −1 (exactly).
The quantity µ
◦(m)
i
is the chemical potential of substance i in its molality standard
state. This standard state is component i in a hypothetical solution with m i equal to
m ◦ (exactly 1 mol kg −1 ) and with Henry’s law in the form of Eq. (6.2-11) valid at this
molality. Again we specify a pressure of exactly 1 bar for this standard state. Since the
standard state is a hypothetical solution, the actual 1-molal solution is not required to
obey Henry’s law.
Exercise 6.11
Show that µ
◦(m)
i
is equal to the chemical potential of substance i in the vapor phase at equilibrium
with a 1.000 mol kg −1 solution if Eq. (6.2-11) is valid at this molality.